Question: You have found the following ages (in years) of all 6 snakes at your local zoo: $ 2,\enspace 21,\enspace 18,\enspace 6,\enspace 26,\enspace 3$ What is the average age of the snakes at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 snakes at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{2 + 21 + 18 + 6 + 26 + 3}{{6}} = {12.7\text{ years old}} $ Find the squared deviations from the mean for each snake. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $2$ years $-10.7$ years $114.49$ years $^2$ $21$ years $8.3$ years $68.89$ years $^2$ $18$ years $5.3$ years $28.09$ years $^2$ $6$ years $-6.7$ years $44.89$ years $^2$ $26$ years $13.3$ years $176.89$ years $^2$ $3$ years $-9.7$ years $94.09$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{114.49} + {68.89} + {28.09} + {44.89} + {176.89} + {94.09}} {{6}} $ $ {\sigma^2} = \dfrac{{527.34}}{{6}} = {87.89\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{87.89\text{ years}^2}} = {9.4\text{ years}} $ The average snake at the zoo is 12.7 years old. There is a standard deviation of 9.4 years.